Integrand size = 23, antiderivative size = 159 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {35 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+a \sin (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{128 \sqrt {2} d}-\frac {35 a^3}{128 d \sqrt {a+a \sin (c+d x)}}+\frac {35 a^2 \sec ^2(c+d x) \sqrt {a+a \sin (c+d x)}}{192 d}+\frac {7 a \sec ^4(c+d x) (a+a \sin (c+d x))^{3/2}}{48 d}+\frac {\sec ^6(c+d x) (a+a \sin (c+d x))^{5/2}}{6 d} \]
7/48*a*sec(d*x+c)^4*(a+a*sin(d*x+c))^(3/2)/d+1/6*sec(d*x+c)^6*(a+a*sin(d*x +c))^(5/2)/d+35/256*a^(5/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^( 1/2))*2^(1/2)/d-35/128*a^3/d/(a+a*sin(d*x+c))^(1/2)+35/192*a^2*sec(d*x+c)^ 2*(a+a*sin(d*x+c))^(1/2)/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.28 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {a^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},4,\frac {1}{2},\frac {1}{2} (1+\sin (c+d x))\right )}{8 d \sqrt {a+a \sin (c+d x)}} \]
-1/8*(a^3*Hypergeometric2F1[-1/2, 4, 1/2, (1 + Sin[c + d*x])/2])/(d*Sqrt[a + a*Sin[c + d*x]])
Time = 0.64 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 3154, 3042, 3154, 3042, 3154, 3042, 3146, 61, 73, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \sec ^7(c+d x) (a \sin (c+d x)+a)^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (c+d x)+a)^{5/2}}{\cos (c+d x)^7}dx\) |
| \(\Big \downarrow \) 3154 |
| \(\displaystyle \frac {7}{12} a \int \sec ^5(c+d x) (\sin (c+d x) a+a)^{3/2}dx+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{12} a \int \frac {(\sin (c+d x) a+a)^{3/2}}{\cos (c+d x)^5}dx+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3154 |
| \(\displaystyle \frac {7}{12} a \left (\frac {5}{8} a \int \sec ^3(c+d x) \sqrt {\sin (c+d x) a+a}dx+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\right )+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{12} a \left (\frac {5}{8} a \int \frac {\sqrt {\sin (c+d x) a+a}}{\cos (c+d x)^3}dx+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\right )+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3154 |
| \(\displaystyle \frac {7}{12} a \left (\frac {5}{8} a \left (\frac {3}{4} a \int \frac {\sec (c+d x)}{\sqrt {\sin (c+d x) a+a}}dx+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\right )+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {7}{12} a \left (\frac {5}{8} a \left (\frac {3}{4} a \int \frac {1}{\cos (c+d x) \sqrt {\sin (c+d x) a+a}}dx+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\right )+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {7}{12} a \left (\frac {5}{8} a \left (\frac {3 a^2 \int \frac {1}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^{3/2}}d(a \sin (c+d x))}{4 d}+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\right )+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {7}{12} a \left (\frac {5}{8} a \left (\frac {3 a^2 \left (\frac {\int \frac {1}{(a-a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}}d(a \sin (c+d x))}{2 a}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}\right )}{4 d}+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\right )+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {7}{12} a \left (\frac {5}{8} a \left (\frac {3 a^2 \left (\frac {\int \frac {1}{2 a-a^2 \sin ^2(c+d x)}d\sqrt {\sin (c+d x) a+a}}{a}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}\right )}{4 d}+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\right )+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {7}{12} a \left (\frac {5}{8} a \left (\frac {3 a^2 \left (\frac {\text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2}}\right )}{\sqrt {2} a^{3/2}}-\frac {1}{a \sqrt {a \sin (c+d x)+a}}\right )}{4 d}+\frac {\sec ^2(c+d x) \sqrt {a \sin (c+d x)+a}}{2 d}\right )+\frac {\sec ^4(c+d x) (a \sin (c+d x)+a)^{3/2}}{4 d}\right )+\frac {\sec ^6(c+d x) (a \sin (c+d x)+a)^{5/2}}{6 d}\) |
(Sec[c + d*x]^6*(a + a*Sin[c + d*x])^(5/2))/(6*d) + (7*a*((Sec[c + d*x]^4* (a + a*Sin[c + d*x])^(3/2))/(4*d) + (5*a*((Sec[c + d*x]^2*Sqrt[a + a*Sin[c + d*x]])/(2*d) + (3*a^2*(ArcTanh[(Sqrt[a]*Sin[c + d*x])/Sqrt[2]]/(Sqrt[2] *a^(3/2)) - 1/(a*Sqrt[a + a*Sin[c + d*x]])))/(4*d)))/8))/12
3.2.38.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p + 1))), x] + Simp[a*((m + p + 1)/(g^2*(p + 1))) Int[(g* Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ [m + 1/2, 2*p]
Time = 0.49 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.71
\[\frac {2 a^{7} \left (-\frac {1}{16 a^{4} \sqrt {a +a \sin \left (d x +c \right )}}-\frac {-\frac {a^{2} \sqrt {a +a \sin \left (d x +c \right )}\, \left (57 \left (\cos ^{2}\left (d x +c \right )\right )+158 \sin \left (d x +c \right )-190\right )}{48 \left (a \sin \left (d x +c \right )-a \right )^{3}}-\frac {35 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{32 \sqrt {a}}}{16 a^{4}}\right )}{d}\]
2*a^7*(-1/16/a^4/(a+a*sin(d*x+c))^(1/2)-1/16/a^4*(-1/48*a^2*(a+a*sin(d*x+c ))^(1/2)*(57*cos(d*x+c)^2+158*sin(d*x+c)-190)/(a*sin(d*x+c)-a)^3-35/32*2^( 1/2)/a^(1/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))))/d
Time = 0.29 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.31 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {105 \, {\left (\sqrt {2} a^{2} \cos \left (d x + c\right )^{4} + 2 \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, \sqrt {2} a^{2} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \log \left (-\frac {a \sin \left (d x + c\right ) + 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 4 \, {\left (245 \, a^{2} \cos \left (d x + c\right )^{2} - 160 \, a^{2} - 7 \, {\left (15 \, a^{2} \cos \left (d x + c\right )^{2} - 32 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{1536 \, {\left (d \cos \left (d x + c\right )^{4} + 2 \, d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )^{2}\right )}} \]
1/1536*(105*(sqrt(2)*a^2*cos(d*x + c)^4 + 2*sqrt(2)*a^2*cos(d*x + c)^2*sin (d*x + c) - 2*sqrt(2)*a^2*cos(d*x + c)^2)*sqrt(a)*log(-(a*sin(d*x + c) + 2 *sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a) + 3*a)/(sin(d*x + c) - 1)) - 4*( 245*a^2*cos(d*x + c)^2 - 160*a^2 - 7*(15*a^2*cos(d*x + c)^2 - 32*a^2)*sin( d*x + c))*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c)^4 + 2*d*cos(d*x + c)^2 *sin(d*x + c) - 2*d*cos(d*x + c)^2)
Timed out. \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\text {Timed out} \]
Time = 0.26 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.16 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {105 \, \sqrt {2} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {a \sin \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {a \sin \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (105 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{3} a^{4} - 560 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a^{5} + 924 \, {\left (a \sin \left (d x + c\right ) + a\right )} a^{6} - 384 \, a^{7}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 6 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a + 12 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2} - 8 \, \sqrt {a \sin \left (d x + c\right ) + a} a^{3}}}{1536 \, a d} \]
-1/1536*(105*sqrt(2)*a^(7/2)*log(-(sqrt(2)*sqrt(a) - sqrt(a*sin(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(a*sin(d*x + c) + a))) + 4*(105*(a*sin(d*x + c ) + a)^3*a^4 - 560*(a*sin(d*x + c) + a)^2*a^5 + 924*(a*sin(d*x + c) + a)*a ^6 - 384*a^7)/((a*sin(d*x + c) + a)^(7/2) - 6*(a*sin(d*x + c) + a)^(5/2)*a + 12*(a*sin(d*x + c) + a)^(3/2)*a^2 - 8*sqrt(a*sin(d*x + c) + a)*a^3))/(a *d)
Time = 0.31 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.91 \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {\sqrt {2} a^{\frac {5}{2}} {\left (\frac {96}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {2 \, {\left (57 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 136 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 87 \, \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}} - 105 \, \log \left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) + 105 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{1536 \, d} \]
-1/1536*sqrt(2)*a^(5/2)*(96/cos(-1/4*pi + 1/2*d*x + 1/2*c) + 2*(57*cos(-1/ 4*pi + 1/2*d*x + 1/2*c)^5 - 136*cos(-1/4*pi + 1/2*d*x + 1/2*c)^3 + 87*cos( -1/4*pi + 1/2*d*x + 1/2*c))/(cos(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^3 - 105 *log(cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1) + 105*log(-cos(-1/4*pi + 1/2*d*x + 1/2*c) + 1))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))/d
Timed out. \[ \int \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\cos \left (c+d\,x\right )}^7} \,d x \]